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प्रश्न
Prove that: `(sec θ - tan θ)/(sec θ + tan θ ) = 1 - 2 sec θ.tan θ + 2 tan^2θ`
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उत्तर
LHS = `(sec θ - tan θ)/(sec θ + tan θ )`
= `(sec θ - tan θ)/(sec θ + tan θ ) xx (sec θ - tan θ)/(sec θ - tan θ )`
= `(sec θ - tan θ)^2/(sec^2θ - tan^2θ )`
= `(sec^2θ + tan^2θ - 2sec θ.tan θ )/1`
= 1 + 2 tan2θ - 2 sec θ. tan θ
= R.H.S.
Hence proved.
संबंधित प्रश्न
Prove the following trigonometric identities.
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Find the value of sin2θ + cos2θ
Solution:
In Δ ABC, ∠ABC = 90°, ∠C = θ°
AB2 + BC2 = `square` .....(Pythagoras theorem)
Divide both sides by AC2
`"AB"^2/"AC"^2 + "BC"^2/"AC"^2 = "AC"^2/"AC"^2`
∴ `("AB"^2/"AC"^2) + ("BC"^2/"AC"^2) = 1`
But `"AB"/"AC" = square and "BC"/"AC" = square`
∴ `sin^2 theta + cos^2 theta = square`
